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Math Help - Represent inverse trigonometric equation as rational integral equation?

  1. #1
    Super Member fardeen_gen's Avatar
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    Represent inverse trigonometric equation as rational integral equation?

    Prove that the equation:
    \cot^{-1}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) = 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}
    can be represented as a rational integral equation in x & y.

    NOTE: No latex for arccot?
    EDIT: @Mush:
    Code:
    \acot
    doesn't work!
    EDIT: Thanks Soroban! Also, \text{arccot} is the only way it seems.
    Last edited by fardeen_gen; April 28th 2009 at 06:49 AM. Reason: Follow up comments.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Prove that the equation:
    \cot^{-1}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) = 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}
    can be represented as a rational integral equation in x & y.

    NOTE: No latex for arccot?
    \acot
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  3. #3
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    Hello, fardeen_gen!

    Prove that the equation: . \text{arccot}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) \;=\; 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}

    can be represented as a rational integral equation in x and y.

    \text{We have: }\;\arctan\underbrace{\left(\frac{\sqrt{1 - x^2 - y^2}}{y}\right)}_{\theta} \:= \:2\cdot\underbrace{\arctan\sqrt{\frac{3 - 4x^2}{4x^2}}}_{\alpha} - \underbrace{\arctan\sqrt{\frac{3 - 4x^2}{x^2}}}_{\beta}

    . . That is: . \theta \;=\;2\alpha - \beta


    Take the tangent of both sides: . \tan\theta \;=\;\tan(2\alpha + \beta)

    . . \tan\theta \;=\;\frac{\tan(2\alpha) + \tan\beta} {1 - \tan(2\alpha)\tan\beta} \;=\;<br />
\frac{\dfrac{2\tan\alpha}{1 - \tan^2\!\alpha} + \tan\beta}{1 - \dfrac{2\tan\alpha}{1 - \tan^2\!\alpha}\tan B}

    . . \tan\theta \;=\;\frac{2\tan\alpha - \tan\beta(1 - \tan^2\!\alpha)} {1-\tan^2\!\alpha - 2\tan\alpha\tan\beta}

    . . \tan\theta - \tan^2\!\alpha\tan\theta - 2\tan\alpha\tan\beta\tan\theta \;=\;2\tan\alpha - \tan\beta + \tan^2\!\alpha\tan\beta .[1]



    Since \tan\theta = \frac{\sqrt{1-x^2-y^2}}{y},\quad \tan\alpha = \frac{\sqrt{3-4x^2}}{2x},\quad \tan\beta = \frac{\sqrt{3-4x^2}}{x}

    substitute into [1]:

    . . \frac{\sqrt{1-x^2-y^2}}{y} \;-\; \frac{3-4x^2}{4x^2} \cdot \frac{\sqrt{1-x^2-y^2}}{y} \;-\; 2 \cdot \frac{\sqrt{3-4x^2}}{2x}\cdot\frac{\sqrt{3-4x^2}}{x}\cdot\frac{\sqrt{1-x^2-y^2}}{y}

    . . . . = \;2\,\frac{\sqrt{3-4x^2}}{2x} - \frac{\sqrt{3-4x^2}}{x} + \frac{3-4x^2}{4x^2}\cdot\frac{\sqrt{3-4x^2}}{x}


    Simplify: . \frac{\sqrt{1-x^2-y^2}}{y} - \frac{5(3-4x^2)\sqrt{1-x^2-y^2}}{4x^2y} \;=\;\frac{(3-4x^2)^{\frac{3}{2}}}{4x^3}


    Multiply by 4x^3y\!:\quad 4x^3\sqrt{1-x^2-y^2} - 5x(3-4x^2)\sqrt{1-x^2-y^2} \;=\;y(3-4x^2)^{\frac{3}{2}}


    Factor and simplify: . 3x(8x^2-5)\sqrt{1-x^2-y^2} \;=\;y(3-4x^2)^{\frac{3}{2}}


    Square both sides: . 9x^2(8x^2-5)^2(1-x^2-y^2) \;=\;y^2(3-4x^2)^3

    . . and we have a polynomial equaton in x and y.

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